What is the Integral of # (1+tanx)^5 * (sec^2(x)) dx #?

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Answer 1 · 2016-12-20T14:53:50

The answer is #=(1+tanx)^6/6+C#

We must calculate #int(1+tanx)^5sec^2xdx#

We do this integral by substitution

Let #u=1+tanx#

#du=sec^2xdx#

#int(1+tanx)^5sec^2xdx#

#=intu^5du=u^6/6#

#=(1+tanx)^6/6+C#